Question

If x is a positive real number and exponents are rational numbers, simplify

${\left(\frac{x^b}{x^c}\right)}^{b+c-a}·{\left(\frac{x^c}{x^a}\right)}^{c+a-b}·{\left(\frac{x^a}{x^b}\right)}^{a+b-c}$

Answer 1

$$\begin{gathered} {\left(\frac{x^b}{x^c}\right)}^{b+c-a}·{\left(\frac{x^c}{x^a}\right)}^{c+a-b}·{\left(\frac{x^a}{x^b}\right)}^{a+b-c} \\ ={\left(x^{b-c}\right)}^{b+c-a}·{\left(x^{c-a}\right)}^{c+a-b}·{\left(x^{a-b}\right)}^{a+b-c} \\ =\left[{\left(x^{b-c}\right)}^b.{\left(x^{b-c}\right)}^{c-a}\right]·{\left(x^{c-a}\right)}^{c+a-b}·\left[{\left(x^{a-b}\right)}^{b-c}.{\left(x^{a-b}\right)}^a\right] \\ =\left[{\left(x^{b-c}\right)}^b.{\left(x^{b-c}\right)}^{c-a}\right]·{\left(x^{c+a-b}\right)}^{c-a}·\left[{\left(x^{a-b}\right)}^{b-c}.{\left(x^{a-b}\right)}^a\right] \\ ={\left(x^{b-c}\right)}^b.\left[{\left(x^{b-c}\right)}^{c-a}·{\left(x^{c+a-b}\right)}^{c-a}\right]·{\left(x^{a-b}\right)}^{b-c}.{\left(x^{a-b}\right)}^a \\ ={\left(x^{b-c}\right)}^b.\left[{\left(x^{b-c+c+a-b}\right)}^{c-a}\right]·{\left(x^{a-b}\right)}^{b-c}.{\left(x^{a-b}\right)}^a \\ ={\left(x^{b-c}\right)}^b.{\left(x^a\right)}^{c-a}·{\left(x^{a-b}\right)}^{b-c}.{\left(x^{a-b}\right)}^a \\ ={\left(x^b\right)}^{b-c}.{\left(x^a\right)}^{c-a}·{\left(x^{a-b}\right)}^{b-c}.{\left(x^{a-b}\right)}^a \\ =\left[{\left(x^b\right)}^{b-c}·{\left(x^{a-b}\right)}^{b-c}\right].{\left(x^a\right)}^{c-a}.{\left(x^{a-b}\right)}^a \\ =\left[{\left(x^{b+a-b}\right)}^{b-c}\right].{\left(x^a\right)}^{c-a}.{\left(x^{a-b}\right)}^a \\ ={\left(x^a\right)}^{b-c}.{\left(x^a\right)}^{c-a}.{\left(x^{a-b}\right)}^a \\ ={\left(x^{b-c}\right)}^a.{\left(x^{c-a}\right)}^a.{\left(x^{a-b}\right)}^a \\ ={\left(x^{b-c+c-a+a-b}\right)}^a \\ =x^0 \\ =1 \end{gathered}$$